How do you solve $x^{2} + 2 x + 8 = 0$ $x^2+2x+8=0$ ?

Question

1262 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-26T15:53:51

The solutions are:
$x = \frac{- 2 + \sqrt{- 28}}{2}$ $x= (-2+sqrt(-28))/2$

$x = \frac{- 2 - \sqrt{- 28}}{2}$ $x=(-2-sqrt(-28))/2$

$x^{2} + 2 x + 8 = 0$ $x^2+2x+8=0$

The equation is of the form $a x^{2} + b x + c = 0$ $color(blue)(ax^2+bx+c=0$ where:

$a = 1, b = 2, c = 8$ $a=1, b=2, c=8$

The Discriminant is given by:

$Delta=b^2-4*a*c$

$= (2)^2-(4*1*8)$

$= 4-32=-28$

The solutions are found using the formula:

$x=(-b+-sqrtDelta)/(2*a)$

$x = ((-2)+-sqrt(-28))/(2*1) = (-2+-sqrt(-28))/2$

The solutions are:
$x= (-2+sqrt(-28))/2$

$x=(-2-sqrt(-28))/2$

How do you solve x^2+2x+8=0x2+2x+8=0x^2+2x+8=0?